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zhannawk [14.2K]
3 years ago
8

Ricky can ride 17 km on his bike in the same length of time ha can walk 9 km.if his riding speed is 4 kph faster than his walkin

g speed how fast does he walk
Mathematics
1 answer:
aivan3 [116]3 years ago
5 0
Distance=speed times time
if they take the same time, we cal the time , t

b=bike speed
w=walking speed

bikedistnace=17=bt
walkdistance=9=wt

b is 4 more than w
b=4+w

we have
17=bt
9=wt
b=4w

ok so
multiply first equation by 9 and 2nd by 17

153=9bt
153=17wt
set equal
9bt=17wt
divide both sides by t
9b=17w
sub b=4+w for b
9(4+w)=17w
distribute
36+9w=17w
minus 9w both sides
36=8w
divide both sides by 8
4.5=w

he walks 4.5kph
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Question 7:

∠L = 124°

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Question 15:

m∠G = 110°

Question 16:

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Question 18:

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Step-by-step explanation:

The angles can be solving using Symmetry.

Question 7.

The sum of interior angles in an isosceles trapezoid is 360°, and because it is an  isosceles trapezoid

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236° + 2 ∠L = 360°

Therefore,

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Question 8.

In a similar fashion,

∠Q+∠T+∠S +∠R = 360°

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∠Q = ∠T

∠Q+∠T + 164° = 360°

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Question 15.

The sum of interior angles of a kite is 360°.

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∠E  = ∠G.

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∠E +∠G + 100° +40° = 360°

2∠E = 140° = 360°

∠E  = 110° = ∠G.

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m∠G = 110°

Question 16.

The angles

∠E = ∠G,

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∠E + ∠G + ∠F +∠H = 360°

∠E + ∠G  + 150 + 90 = 360°

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Question 17.

The shape is a kite; therefore,

∠H = ∠F = 110°

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220° + 60° + ∠G = 360°,

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Question 18.

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∠F = ∠H  = 90°

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we know the cosine is positive, and we know the opposite side is -1, or negative, the only happens in the IV quadrant, so θ is in the IV quadrant, now

\bf csc(\theta)=-6\implies csc(\theta)=\cfrac{\stackrel{hypotenuse}{6}}{\stackrel{opposite}{-1}}\impliedby \textit{let's find the \underline{adjacent side}}
\\\\\\
\textit{using the pythagorean theorem}\\\\
c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a
\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}
\\\\\\
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% tangent
tan(\theta)=\cfrac{opposite}{adjacent}
\qquad \qquad 
% cotangent
cot(\theta)=\cfrac{adjacent}{opposite}
\\\\\\
% cosecant
csc(\theta)=\cfrac{hypotenuse}{opposite}
\qquad \qquad 
% secant
sec(\theta)=\cfrac{hypotenuse}{adjacent}

therefore, let's just plug that on the remaining ones,

\bf sin(\theta)=\cfrac{-1}{6}
\qquad\qquad 
cos(\theta)=\cfrac{\sqrt{35}}{6}
\\\\\\
% tangent
tan(\theta)=\cfrac{-1}{\sqrt{35}}
\qquad \qquad 
% cotangent
cot(\theta)=\cfrac{\sqrt{35}}{1}
\\\\\\
sec(\theta)=\cfrac{6}{\sqrt{35}}

now, let's rationalize the denominator on tangent and secant,

\bf tan(\theta)=\cfrac{-1}{\sqrt{35}}\implies \cfrac{-1}{\sqrt{35}}\cdot \cfrac{\sqrt{35}}{\sqrt{35}}\implies \cfrac{-\sqrt{35}}{(\sqrt{35})^2}\implies -\cfrac{\sqrt{35}}{35}
\\\\\\
sec(\theta)=\cfrac{6}{\sqrt{35}}\implies \cfrac{6}{\sqrt{35}}\cdot \cfrac{\sqrt{35}}{\sqrt{35}}\implies \cfrac{6\sqrt{35}}{(\sqrt{35})^2}\implies \cfrac{6\sqrt{35}}{35}
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3 years ago
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